- 6 t x x = X x = X−54 x = 12 L L L L L L LL Hare Tortoise
Let the times of the first and second meetings be T1and T2. Then
vT1= 12, V (T1−12) =12, vT2= (X−54) , V (T2− 1) = X +54.
The first pair and the second pair of equations give, respectively, 1 v − 1 V = 1 , X−54 v = X +54 V + 1 ,
and hence (first eliminating v):
V = 10
4X− 9, v = 10 4X + 1. The total distance travelled by the hare in1
2hour is 2X, so 2X = 10 4X− 9× 1 2, i.e. 8X 2− 18X − 5 = 0 .
Factorising the quadratic gives two roots−1 4and
5
2, and we obviously need the positive root.
The speed of the tortoise is given by
v = 10
4X + 1 = 10 11 so the time taken to travel5
2 kilometres is 11
4 hours.
Post-mortem
It is interesting to see how quickly a mechanics question can change from mechanics to something else: algebra, calculus or, as in this case, coordinate geometry. As soon as the picture is drawn, the equations of the lines can be written down and everything follows as if it were a geometry problem.
Sometimes the change happens even more quickly. There is a well-known problem about a monk as- cending a hill one day and descending the next day starting at the same time on each day. The problem is to show that there is a time at which the monk was at the same height on both days. You are not even given the speed at which the monk climbs. But if you draw (or just imagine drawing) height-time graphs of the ascent and descent, and superimpose them, you are done.
1997 Paper I
A single stream of cars, each of width a and exactly in line, is passing along a straight road of breadth b with speed V . The distance between successive cars (i.e. the distance between back of one car and the front of the following car) is c.
A chicken crosses the road in safety at a constant speed u in a straight line making an angle θ with the direction of traffic. Show that
u⩾ V a
c sin θ + a cos θ. (∗)
Show also that if the chicken chooses θ and u so that she crosses the road at the least possible (constant) speed, she crosses in time
b V (c a+ a c ) .
Comments
I like this question because it relates to (an idealised version of) a situation we have probably all thought about. Once you have visualised it, there are no great difficulties. As usual, you have to be careful with the inequalities, though it turns out here that there is no danger of dividing by a negative quantity.
Solution to problem 54
The easiest way to think about this problem is to consider the cars to be stationary and the velocity of the chicken to be (u cos θ− V, u sin θ). Then the diagrams are very easy to visualise.
Let t be the time taken to cross the distance a in which the chicken is at risk. Then a = ut sin θ.
For safety, the chicken must choose ut cos θ + c⩾ V t: equality here occurs when the chicken starts at the near-side rear of one car and just avoids being hit by the far-side front of the next car.
Eliminating t from these two equations gives the required inequality:
ut cos θ⩾ V t − c
=⇒ (u cos θ − V )t ⩾ −c =⇒ (u cos θ − V ) a
u sin θ ⩾ −c
=⇒ au cos θ − aV ⩾ −cu sin θ =⇒ u(c sin θ + a cos θ) ⩾ aV which is the required result.
For a given value of θ, the minimum speed satisfies
u(c sin θ + a cos θ) = aV.
The smallest value of this u is therefore obtained when c sin θ + a cos θ is largest. This can be found by calculus (regard it as a function of θ and differentiate: the maximum occurs when tan θ = c/a) or by trigonometry:
c sin θ + a cos θ =√a2+ c2cos (θ− arctan(c/a))
so the maximum value is√a2+ c2and it occurs when tan θ = c/a.
The time of crossing is
b u sin θ = b(c sin θ + a cos θ) V a sin θ = b(c + a cot θ) V a = b(c + a2/c) V a .
Post-mortem
There is another inequality besides (∗) that you might have noticed. If u cos θ > V (so the chicken moves faster than the cars — a bit unlikely unless the chicken is trying to cross the M25), the chicken should start her run at the front nearside of a car and must not collide with the car ahead. This requires (u cos θ− V )t ⩽ c, so
u(−c sin θ + a cos θ) ⩽ aV.
If (−c sin θ + a cos θ) < 0, this places no constraint on u. But if (−c sin θ + a cos θ) > 0, then
u⩽ aV
−c sin θ + a cos θ.
In both cases, the inequality (∗) overleaf does not apply. This is clearly not the situation envisaged by the examiners, and probably not by any of the candidates either, but still it should have been catered for in the wording of the question.
1998 Paper I
Hank’s Gold Mine has a very long vertical shaft of height l. A light chain of length l passes over a small smooth light fixed pulley at the top of the shaft. To one end of the chain is attached a bucket A of negligible mass and to the other a bucket B of mass m.
The system is used to raise ore from the mine as follows. When bucket A is at the top it is filled with mass 2m of water and bucket B is filled with mass λm of ore, where 0 < λ < 1. The buckets are then released, so that bucket A descends and bucket B ascends. When bucket B reaches the top both buckets are emptied and released, so that bucket B descends and bucket A ascends. The time to fill and empty the buckets is negligible. Find the time taken from the moment bucket
Ais released at the top until the first time it reaches the top again.
This process goes on for a very long time. Show that, if the greatest amount of ore is to be raised in that time, then λ must satisfy the condition f′(λ) = 0where
f(λ) = λ(1− λ)
1/2
(1− λ)1/2+ (3 + λ)1/2.
Comments
One way of working out the acceleration of a system of two masses connected by a light string passing over a pulley is to write down the equation of motion of each mass, bearing in mind that the force due to tension will be the same for each mass (it cannot vary along the string, because then the acceleration of some portion of themasslessstring would be infinite). Then you eliminate the tension.
Alternatively, you can use the equation of motion of the system of two joined masses. The system has inertial mass equal to the sum of the masses (because both masses must accelerate equally) but gravitational mass equal to the difference of the masses (because the gravitational force on one mass cancels, partially, the gravitational force on the other), so the equation of motion is just (Newton’s law of motion)
Solution to problem 55
When bucket A ascends, the acceleration is g.
For bucket A’s downward journey, at acceleration a, the equations of motion for bucket A and bucket
B, respectively, are
−T + 2mg = 2ma, T − (1 + λ)mg = (1 + λ)ma,
where T is the tension in the rope. Eliminating T gives so a = 1− λ 3 + λg. The time of descent (using l = 1
2at
2) is√2l/aand the time of ascent is√2l/g. The total time required
for one complete cycle is therefore √ 2l g ( 1 + √ 3 + λ 1− λ ) .
Call this t. The number of round trips in a long time tlongis tlong/tso the amount of ore lifted in time
tlongis λmtlong/t.
To maximise this, we have to maximise λ/t with respect to λ, and λ/t is exactly the f (λ) given. Note that f (0) = 0 (which makes sense because no ore is raised if λ = 0), and f (1) = 0 (which also makes sense because the buckets don’t move on the raising stage if λ = 1). That means the greatest value of f (λ)must occur at a value of λ in the range 0 < λ < 1 at which f′(λ) = 0.
Post-mortem
You may have wondered why, in the last part, the question says that the process goes on for a very long time. The reason for this is that when you maximise what I have called λmtlong/tthe result may
not correspond to a complete number of cycles. If you stop in mid-cycle, you raise no ore from that cycle so a calculus maximisation of a continuous function is not the right method. However, if the process continues for many cycles, the contribution from the last cycle becomes negligible, and a calculus maximisation becomes appropriate.
You may also wonder why you were not asked to find the maximising value of λ. If you are feeling exceptionally energetic, you could try to solve f′(λ) = 0. This will eventually lead you to the rather discouraging quartic equation
λ4+ 4λ3+ 2λ2− 8λ + 3 = 0 .
You will have done a lot of tedious work to obtain an equation that cannot be solved without either a formula for the solutions of a general quartic equation or a computer. There are in fact two real solutions of this equation, λ = 0.528 and λ = 0.704.
1987 Specimen Paper II
A chocolate orange consists of a sphere of delicious smooth uniform chocolate of mass M and radius a, sliced into segments by planes through a fixed axis. It stands on a horizontal table with this axis vertical and it is held together by a narrow ribbon round its equator. Show that the tension in the ribbon is at least 3
32M g.
[You may assume that the centre of mass of a segment of angle 2θ is at distance 3πa sin θ 16θ from the axis.]
Comments
This question can be done by the usual methods (resolving forces and taking moments about a suitably chosen point). Since the chocolate is smooth, there is no friction. The ribbon may be elastic, so it could be tighter than is needed just to keep the orange together. At the lowest tension possible the orange is on the point of falling apart, so there are no forces between the faces of the segments, except at the point of contact with the table.
The ribbon is said to be thin, but actually this has to be interpreted as massless as well as thin; a ribbon with mass would simply drop off the equator of the orange.
Notice that the question does not say that the segments are all the same size. That suggests that we should look at just one segment, expecting to find that the tension required for that segment is indepen- dent of the angle of the segment.
When I set this question originally on a 1985 examination paper (pre-dating STEP, which started in 1987), I gave the wrong formula for the distance of the centre of mass of the segment, and the answer for the tension was also incorrect, but consistent with the incorrect formula. Not surprisingly, no one pointed it out, either at the time or afterwards. With a bit of luck, it is correct now.
I sent a copy of the examination paper to a well-known manufacturer of high quality chocolate con- fection31and was rewarded with a puzzled letter and a parcel; not as large a parcel as I had hoped for
(perhaps they spotted the incorrect formula), but better than nothing. It seemed worth including the question in this book in case they wanted an opportunity to make amends . . . .
31Now located in Poland. Originally, the firm made chocolate apples which, being apple-shaped, would not have worked well
for this question (the centre of mass of an orange segment was hard enough for me). The chocolate lemon, introduced in 1974, would not have been good either but luckily it was discontinued rapidly after it turned out that no one wanted to eat it.